Prove if f(p) = 0 where abs(f) is continuous at p, then f is continuous at p. Note: we have only learned continuity, discontinuty, and epsilon-delta proofs (no limits)!

Answer 1

epsilon-delta proofs use limits.

abs(f) is continuous if f is continuous, because

|f| = f if f>=0
|f| = -f is f<0

So, if the limit L exsists for f(x), it also exists for |f(x)|.